Hello Moo! I tried on the question with your advise and this is what I got so far! I am not sure if I am on the right track, but correct me if I am wrong! (:

For part (i)

According to the Chebyshev's inequality, if X has mean $\displaystyle \mu$ and variance, $\displaystyle \sigma^2$, then,$\displaystyle \mathbb P (|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}$

From the question,$\displaystyle \mathbb P(|\mathbb Z_n - p| \geq \epsilon) \leq \frac{1}{4n \epsilon^2}$

$\displaystyle \Rightarrow \mu = p

k = 2 \sqrt n \epsilon

\sigma = \frac{2}{2\sqrt n}$

$\displaystyle \sigma^2 = E[(z_n - p)^2]$

$\displaystyle = \int^\infty_{-\infty} (z_n - p)^2 f(x) dx$

$\displaystyle \geq \int_{|z_n - p| \geq \epsilon} (z_n - p)^2 f(x) dx$

$\displaystyle \geq (2 \sqrt n \epsilon)^2(\sigma^2) \int_{|z_n - p| \geq \epsilon} f(x) dx$

$\displaystyle \geq 4n\epsilon^2(\sigma^2) \int_{|z_n - p| \geq \epsilon} f(x)$

$\displaystyle \geq 2n \epsilon^2 (\sigma^2) \mathbb P(Z_n - p) \geq \epsilon)$

By dividing both sides with $\displaystyle \sigma^2$,

$\displaystyle \Rightarrow \mathbb P(|Z_n - p| \geq \epsilon) \leq \frac{1}{4n\epsilon^2}$

For part (ii),$\displaystyle \mathbb P(|Z_n - P) \leq \epsilon)$

$\displaystyle \Rightarrow \mathbb P(|Z_n - p| \leq \epsilon) \geq 1 - \frac{1}{4n\epsilon^2}$

$\displaystyle \Rightarrow 1 - \frac{1}{4n\epsilon^2} \geq 0.95$

$\displaystyle \Rightarrow \frac{1}{4n\epsilon^2} \leq 0.05$

$\displaystyle \Rightarrow 4n\epsilon^2 \geq 20$

By estimating p within 0.1 means we let $\displaystyle \epsilon = 0.01$,$\displaystyle \Rightarrow 4n(0.01)^2 \geq 20$

$\displaystyle \Rightarrow 4n \geq 200000$

$\displaystyle \Rightarrow n \geq 50000$

$\displaystyle \Rightarrow n = 50001$.

Am I doing alright so far?

Hm, Central Limit Theorem says ...

Let $\displaystyle X_1, X_2, ...$ be indepedent, identically distributed random variables with $\displaystyle E(X_i) = \mu$ and $\displaystyle V(X_i) = \sigma^2$, and let $\displaystyle S_n = X_1 + X_2 + ... + X_n$. Then,$\displaystyle Z_n = \frac {S_n - n\mu}{\sigma \sqrt n} \rightarrow N(0,1),$ as n $\displaystyle \rightarrow \infty$.

Not really sure how to proceed from here!